Ohm's law
$$
1: V(t) = I(t)R
$$
Instantaneous power dissipation is product of voltage and current
$$
2: P(t) = V(t)I(t)\\
$$
Substitute 1 into 2 to get instantaneous power through a resistor in terms of voltage or current:
$$
3: P(t) = I^2(t)R = \frac{V^2(t)}{R}\\
$$
Average power is definitionally the integral of instantaneous power over a period, divided by that period. Substitute 3 into that to get average power in terms of voltage and current.
$$
4: P_{avg}=\frac{\int_0^T{P(t)dt}}{T}=\frac{R\int_0^T{I^2(t)dt}}{T}=\frac{\int_0^T{V^2(t)dt}}{RT}\\
$$
Definition of RMS current
$$
5: I_{RMS}=\sqrt{\frac{\int_0^T{I^2(t)dt}}{T}}\\
$$
Square both sides
$$
6: I_{RMS}^2 =\frac{\int_0^T{I^2(t)dt}}{T}\\
$$
Multiply by R to find equation 4 for average power
$$
7: I_{RMS}^2R =\frac{R\int_0^T{I^2(t)dt}}{T}=P_{avg}\\
$$
Definition of RMS voltage
$$
8: V_{RMS}=\sqrt{\frac{\int_0^T{V^2(t)dt}}{T}}\\
$$
Square both sides
$$
9: V_{RMS}^2=\frac{\int_0^T{V^2(t)dt}}{T}\\
$$
Divide by R to find equation 4 for average power
$$
10: \frac{V_{RMS}^2}{R}=\frac{\int_0^T{V^2(t)dt}}{RT}=P_{avg}\\
$$
Multiply expressions 7 and 10 for average power
$$
11: P_{avg}^2=V_{RMS}^2I_{RMS}^2\\
$$
Square root of both sides
$$
12: P_{avg} = V_{RMS}I_{RMS}\\
$$
Q.E.D.
Why does reactive power influence the voltage? Suppose you have a (weak) power system with a large reactive load. If you suddenly disconnect the load, you would experience a peak in the voltage.
First, we need to define what exactly is being asked. Now that you have stated this is regarding a utility-scale power system, not the output of a opamp or something, we know what "reactive power" means. This is a shortcut used in the electric power industry. Ideally the load on the system would be resistive, but in reality is is partially inductive. They separate this load into the pure resistive and pure inductive components and refer to what is delivered to the resistance as "real power" and what is delivered to the inductance as "reactive power".
This gives rise to some interesting things, like that a capacitor accross a transmission line is a reative power generator. Yes, that sounds funny, but if you follow the definition of reactive power above, this is all consistant and no physics is violated. In fact, capacitors are sometimes used to "generate" reactive power.
The actual current coming out of a generator is lagging the voltage by a small phase angle. Instead of thinking of this as a magnitude and phase angle, it is thought of as two separate components with separate magnitudes, one at 0 phase and the other lagging at 90° phase. The former is the current that causes real power and the latter reactive power. The two ways of describing the overall current with respect to the voltage are mathematically equivalent (each can be unambiguously converted to the other).
So the question comes down to why does generator current that is lagging the voltage by 90° cause the voltage to go down? I think there are two answers to this.
First, any current, regardless of phase, still causes a voltage drop accross the inevitable resistance in the system. This current crosses 0 at the peak of the voltage, so you might say it shouldn't effect the voltage peak. However, the current is negative right before the voltage peak. This can actually cause a little higher apparent (after the voltage drop on the series resistance) voltage peak immediately before the open-circuit voltage peak. Put another way, due to non-zero source resistance, the apparent output voltage has a different peak in a different place than the open-circuit voltage does.
I think the real answer has to do with unstated assumptions built into the question, which is a control system around the generator. What you are really seeing the reaction to by removing reactive load is not that of the bare generator, but that of the generator with its control system compensating for the change in load. Again, the inevitable resistance in the system times the reactive current causes real losses. Note that some of that "resistance" may not be direct electrical resistance, but mechanical issues projected to the electrical system. Those real losses are going to add to the real load on the generator, so removing the reactive load still relieves some real load.
This mechanism gets more substantial the wider the "system" is that is producing the reactive power. If the system includes a transmission line, then the reactive current is still causing real I2R losses in the transmission line, which cause a real load on the generator.
Best Answer
Such explanations are wrong. Sources of such information should probably be avoided.
"Reactive power" is a common term for reactive volt-amperes (VARs). Reactive volt-amperes have one very important use, they provide the magnetic fields for every electromagnetic mechanism. Induction motors represent most of the requirement for VARs, but transformers, AC solenoids and relays also require VARs.
Since VARs represent energy that is continually circulated back and forth between the source and load, the only power consumed is due to losses. That is a small amount and is measured as watts when it occurs on the consumer side of the meter. On the utility side of the meter, the losses are a concern for the utility. The utility is also concerned with the use of transmission capacity to carry current that is not delivering power.
Only synchronous generators and capacitors can provide VARs to magnetic loads. If capacitors are connected near the magnetic loads, that avoids the associated transmission losses and frees up generating, transmission and distribution capacity to provide more real power. However if too much capacitance is added to supply VARs, the utility must accept the extra VARs. Only synchronous generators and inductors can accept VARs from excess capacitance. Since utilities are not expecting excess capacitance, they are not prepared to add inductance. The synchronous generators may have difficulty with that kind of load. The result would be excess voltage.
Since the statement is not entirely correct, there is no proof. Understanding what can happen requires studying the "V curves" for synchronous generators. It is also necessary to understand the various limits on the safe operating conditions for synchronous generators and transmission system component.
More re increased voltage
In order to overcome voltage drops in the transmission system, the voltage at the generator must be higher (on a percentage basis) than the voltage seen by a customer. If the customer had been "consuming VARs," adding capacitors would reduce the current and voltage drop in the transmission lines. That would tend to increase the voltage at the meter, but the utility would adjust the generator to keep the customer voltage within limits. If the customer begins to supply vars, that would tend to make the voltage higher at the customer location. To some extent, VARs supplied by one customer may be absorbed by other customers. That would tend to make that customer's voltage higher than the voltage seen by other consumers.